Question

Two long conductors are arranged as shown above to form overlapping
cylinders, each of raidus r, whose centers are separated by a distance d. Current
of density J flows into the plane of the page along the shaded part of one
conductor and an equal current flows out of the plane of the page along the
shaded portion of the other, as shown. What are the
magnitude and direction of the magnetic field at point A?

• A. $\left({\mu }_0/2\pi \right)\pi dJ$, in the $+y$-direction
• B. $\left({\mu }_0/2\pi \right)d^2/r$, in the $+y$-direction
• C. $\left({\mu }_0/2\pi \right)4d^{2}J/r$, in the-y-direction
• D. $\left({\mu }_0/2\pi \right)J{r}^2/d$, in the-y-direction

Answer 1

Answer: (A)

$\left({\mu }_0/2\pi \right)\pi dJ$, in the $+y$-direction

Given:

Radius of the cylinders, $r$

Distance between their centers $=d$

For the left hand cylinder,

Current density $=\overrightarrow{J}$

For right hand cylinder,

Current density $=-\overrightarrow{J}$

Applying ampere's circuital law to the left cylinder,

${\overrightarrow{B}}_1.dl={\mu }_o\overrightarrow{I}$

${\overrightarrow{B}}_1.2\pi r={\mu }_o\overrightarrow{J}\times \pi r^2$

${\overrightarrow{B}}_1=\frac{{\mu }_o\overrightarrow{J}\times \overrightarrow{r}}{2}$

Similarly, for the right cylinder,

Since, the distance between the centres of the cylinders is $d$,

The Ampere's loop will taken with radius $(d-r)$

Thus,

${\overrightarrow{B}}_2=\frac{{\mu }_o\overrightarrow{J}\times (\overrightarrow{d}-\overrightarrow{r})}{2}$

Net field at the intersecting part $={\overrightarrow{B}}_1+{\overrightarrow{B}}_2=\frac{{\mu }_o\overrightarrow{J}\times \overrightarrow{d}}{2}$

Here, the current density have unit vector along +z-axis and distance vector have direction +x-axis, thus the resultant magnetic field will have the direction of +y-axis.